`1)n\inN^{**}`
`2n-3\vdots n+1`
`=>2n+2-5\vdots n+1`
`=>5\vdots n+1`
`=>n+1\in Ư(5)={+-1,+-5}`
`=>n\in{0;-2;4;-6}`
Mà `n\in N^{**}`
`=>n=4`
Vậy với `n=4` thì `2n-3\vdots n+1`.
`9)A=1/(|x+2017|+|x-2|)`
Áp dụng tính chất:`|P|>=P,|P|>=-P`
`=>{(|x+2017|>=x+2017),(|x-2|>=2-x):}`
`=>|x+2017|+|x-2|>=x+2017+2-x=2019>0`
`=>A<=1/2019`
Dấu "=" xảy ra khi `{(x+2017>=0),(x-2<=0):}`
`<=>-2017<=x<=2`
Vậy `Max_A=1/2019<=>-2017<=x<=2.`
